# Showing that $l^p(\mathbb{N})^* \cong l^q(\mathbb{N})$

I'm reading functional analysis in the summer, and have come to this exercise, asking to show that the two spaces $l^p(\mathbb{N})^*,l^q(\mathbb{N})$ are isomorphic, that is, by showing that every $l \in l^p(\mathbb{N})^*$ can be written as $l_y(x)=\sum y_nx_n$ for some $y$ in $l^q(\mathbb N)$.

The exercise has a hint. Paraphrased: "To see $y \in l^q(\mathbb N)$ consider $x^N$ defined such that $x_ny_n=|y_n|^q$ for $n \leq N$ and $x_n=0$ for $n > N$. Now look at $|l(x^N)| \leq ||l|| ||x^N||_p$."

I can't say I understand the first part of the hint. To prove the statement I need to find a $y$ such that $l=l_y$ for some $y$. How then can I define $x$ in terms of $y$ when it is $y$ I'm supposed to find. Isn't there something circular going on?

The exercise is found on page 68 in Gerald Teschls notes at http://www.mat.univie.ac.at/~gerald/ftp/book-fa/index.html

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I think you're supposed to define $y$ by $y_n = l(e_n)$ ($e_n$ has a $1$ as its $n$-th entry and zeros everywhere else). Is that the problem? – Dylan Moreland Aug 6 '11 at 22:32
From your name and the fact that you're reading notes from Austria I assume that you speak German. If so, I'd recommend Werner's book Funktionalanalysis because it looks like it could be a very good accompanying book to Teschls notes (I only looked at the TOC of the latter). – t.b. Aug 7 '11 at 1:10
@Theo: Nice suggestion! And it is quite inexpensive. – Jonas Teuwen Aug 7 '11 at 8:30
@Jonas: It's a really nice if not outstanding book. The only criticism I have is that it is a bit too detailed. It would be great if somebody would translate it to English, as it is way better than many of the standard books. The notes and comments sections at the end of each chapter are extremely nice and contain lots of references for further study. If you're lucky, you might even be able to download it using your university account. See also the comments by vonjd here for some further links: math.stackexchange.com/q/38460 – t.b. Aug 7 '11 at 8:42
@Theo: I'm Norwegian, but I read a bit German. I have never read mathematics in German, so I do not know if my knowledge of the language is sufficient. – Fredrik Meyer Aug 7 '11 at 22:03

We know that the $(e_n)$ with a $1$ at the $n$-th position and $0$s elsewhere is a Schauder basis for $\ell^p$ (which has some nice alternative equivalent definitions, I recommend Topics in Banach Space Theory by Albiac and Kalton as a nice reference about this.

So, every $x \in \ell^p$ has a unique representation by

$$x = \sum_{k = 1}^\infty y_k e_k.$$

Now consider $l \in \ell^q$. Because $l$ is bounded we also have that

$$l(x) = \sum_{k = 1}^\infty y_k l(e_k).$$

Now set $z_k = f(e_k)$. Consider the following $x_n = (y_k^{(n)})$ where

$$y_k^{(n)} = \begin{cases} \frac{|z_k|^q}{z_k} &\text{when k \leq n and z_k \neq 0,}\\ 0 &\text{otherwise.} \end{cases}.$$

We have that \begin{align}l(x_n) &= \sum_{k = 1}^\infty y_k^{(n)} z_k\\ &= \sum_{k = 1}^n |z_k|^q\\ &\leq \|l\|\|x_n\|\\ &= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\ &= \|l\| \left ( \sum |x_k^{(n)}|^p \right )^{\frac1p}\\ &= \|l\| \left ( \sum |z_k|^q \right )^{\frac1q}. \end{align}

Hence we have that

$$\sum |z_k|^q = \|l\| \left (\sum |z_k|^q \right )^{\frac1q}.$$

Now we divide and get

$$\left ( \sum_{k = 1}^n |z_k|^q \right )^{\frac1q} \leq \|l\|.$$

Take the limit to obtain $$\left ( \sum_{k \geq 1} |z_k|^q \right )^{\frac1q} \leq \|l\|.$$

We conclude that $(z_k) \in \ell^q$.

So, now you could try doing the same for $L^p(\mathbf R^d)$ with a $\sigma$-finite measure. A small hint: Using the $\sigma$-finiteness you can reduce to the finite measure case.

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Thanks for the pointer to Albiac-Kalton, I didn't know that book before, looks really good (I should read more stuff by Kalton anyway). Since we had Werner mentioned already, he wrote a nice summary on Kalton's work recently: arxiv.org/abs/1103.3153 – t.b. Aug 7 '11 at 10:10
@Theo: Too bad Nigel Kalton passed away last year. The same happened to Hans Duistermaat... That book is one of the best that I've seen on this subject. Thanks for the link. – Jonas Teuwen Aug 7 '11 at 10:19
Thanks for the answer! I'm not sure however if I understand every step. Going to the last inequality, I only get $l(x_n) ||z||_q \leq ||l|| \sum^n |z_k|^q$. I don't see why we define $x_n$ as a sequence - where do we use the finiteness this defintion gives? – Fredrik Meyer Aug 7 '11 at 15:39
@Fredrik: I have added some calculations, I hope I didn't mess something up. – Jonas Teuwen Aug 7 '11 at 16:56
Does the prove work for $L^p(R^d)$ ? How do you define the analogue of $(z_k)$ in this case ? I only know a similar proof when $p \in [1,2]$, because you have the embedding $L^2([0,1]^d) \rightarrow L^p([0,1]^d)$ which allows you to construct the "$(z_k)$". – Auguste Hoang Duc Aug 7 '11 at 16:57